2D Computer Graphicsw3.impa.br/~diego/teaching/2020.0/slides-4.pdf · Curvemodelingbysplines...
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2D Computer Graphics Diego Nehab Summer 2020 IMPA 1
Transcript of 2D Computer Graphicsw3.impa.br/~diego/teaching/2020.0/slides-4.pdf · Curvemodelingbysplines...
2D Computer Graphics
Diego Nehab
Summer 2020
IMPA
1
Bézier curves
Curve modeling by splines
2
Curve modeling by splines
Thin strip of wood used in building construction
Anchored in place by lead weights called ducks
Physical process
Interpolating, smooth, energy minimizing 3
No local control 7
3
Curve modeling by splines
Thin strip of wood used in building construction
Anchored in place by lead weights called ducks
Physical process
Interpolating, smooth, energy minimizing 3
No local control 7
3
Curve modeling by splines
Thin strip of wood used in building construction
Anchored in place by lead weights called ducks
Physical process
Interpolating, smooth, energy minimizing 3
No local control 7
3
Curve modeling by splines
Thin strip of wood used in building construction
Anchored in place by lead weights called ducks
Physical process
Interpolating, smooth, energy minimizing 3
No local control 7
3
Curve modeling by splines
Thin strip of wood used in building construction
Anchored in place by lead weights called ducks
Physical process
Interpolating, smooth, energy minimizing 3
No local control 7
3
Burmester curve
4
Lagrangian interpolation
Computational process
k+ 1 vertices {p0, . . . ,pk} define a curve
γ(t) =k∑
i=0
pi
∏j 6=i(t − j)∏j 6=i(i− j)
, t ∈ [0, k]
Similar issues
5
B-splines
Define a family of generating functions βn recursively
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
βn = βn−1 ∗ β0, n ∈ N
Notation for convolution
h = f ∗ g ⇔ h(t) =
∫ ∞
−∞f (u)g(u− t)dt
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
6
B-splines
Define a family of generating functions βn recursively
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
βn = βn−1 ∗ β0, n ∈ N
Notation for convolution
h = f ∗ g ⇔ h(t) =
∫ ∞
−∞f (u)g(u− t)dt
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
6
B-splines
Define a family of generating functions βn recursively
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
βn = βn−1 ∗ β0, n ∈ N
Notation for convolution
h = f ∗ g ⇔ h(t) =
∫ ∞
−∞f (u)g(u− t)dt
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
6
B-splines
Define a family of generating functions βn recursively
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
βn = βn−1 ∗ β0, n ∈ N
Notation for convolution
h = f ∗ g ⇔ h(t) =
∫ ∞
−∞f (u)g(u− t)dt
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
6
B-splines
Define a family of generating functions βn recursively
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
βn = βn−1 ∗ β0, n ∈ N
Notation for convolution
h = f ∗ g ⇔ h(t) =
∫ ∞
−∞f (u)g(u− t)dt
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
-4 -2 2 4-0.2
0.2
0.4
0.6
0.8
1.
1.2
6
B-splines
Examples
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
β1(t) =
1+ t, −1 ≤ t < 0
1− t, 0 ≤ t < 1
0, otherwise
β2(t) =
18(3+ 2t)2, − 3
2 ≤ t < − 12
14(3− 4t2), − 1
2 ≤ t < 12
18(9− 12t + 4t2), 1
2 ≤ t < 22
0, otherwise
7
B-splines
Examples
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
β1(t) =
1+ t, −1 ≤ t < 0
1− t, 0 ≤ t < 1
0, otherwise
β2(t) =
18(3+ 2t)2, − 3
2 ≤ t < − 12
14(3− 4t2), − 1
2 ≤ t < 12
18(9− 12t + 4t2), 1
2 ≤ t < 22
0, otherwise
7
B-splines
Examples
β0(t) =
1, − 12 ≤ t < 1
2
0, otherwise
β1(t) =
1+ t, −1 ≤ t < 0
1− t, 0 ≤ t < 1
0, otherwise
β2(t) =
18(3+ 2t)2, − 3
2 ≤ t < − 12
14(3− 4t2), − 1
2 ≤ t < 12
18(9− 12t + 4t2), 1
2 ≤ t < 22
0, otherwise
7
B-splines
k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve
γ(t) =k∑
i=0
βn(t − i)pi, t ∈ [0, k]
Local control 3
Differentiable n times everywhere 3
Non-interpolating 7
• Interpolation requires solving a banded linear system 3
Many, many interesting properties
8
B-splines
k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve
γ(t) =k∑
i=0
βn(t − i)pi, t ∈ [0, k]
Local control 3
Differentiable n times everywhere 3
Non-interpolating 7
• Interpolation requires solving a banded linear system 3
Many, many interesting properties
8
B-splines
k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve
γ(t) =k∑
i=0
βn(t − i)pi, t ∈ [0, k]
Local control 3
Differentiable n times everywhere 3
Non-interpolating 7
• Interpolation requires solving a banded linear system 3
Many, many interesting properties
8
B-splines
k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve
γ(t) =k∑
i=0
βn(t − i)pi, t ∈ [0, k]
Local control 3
Differentiable n times everywhere 3
Non-interpolating 7
• Interpolation requires solving a banded linear system 3
Many, many interesting properties
8
B-splines
k+ 1 vertices {p0, . . . ,pk} and generating function βn define a curve
γ(t) =k∑
i=0
βn(t − i)pi, t ∈ [0, k]
Local control 3
Differentiable n times everywhere 3
Non-interpolating 7
• Interpolation requires solving a banded linear system 3
Many, many interesting properties
8
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Polylines
k+ 1 vertices {p0, . . . ,pk} define k segments{γ0(t), . . . , γk−1(t)
}, t ∈ [0, 1]
Each segment defined by linear interpolation
γi(t) = (1− t)pi + t pi+1, i ∈ {0, . . . , k− 1}
Local control 3
Interpolates all control points 3
Not differentiable at interpolated points 7
9
Bézier curves
Generalization of linear interpolation
kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γ
nn(t), . . . , γ
n(k−1)n(t)
}, t ∈ [0, 1]
Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈
{nj, . . . ,n(j+ 1)
},
γmi (t) = (1− t) γm−1i
(t) + t γm−1i+1
(t), i ∈{nj, . . . ,n(j+ 1)−m
}De Casteljau algorithm
Geometric interpretation
10
Bézier curves
Generalization of linear interpolation
kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γ
nn(t), . . . , γ
n(k−1)n(t)
}, t ∈ [0, 1]
Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈
{nj, . . . ,n(j+ 1)
},
γmi (t) = (1− t) γm−1i
(t) + t γm−1i+1
(t), i ∈{nj, . . . ,n(j+ 1)−m
}De Casteljau algorithm
Geometric interpretation
10
Bézier curves
Generalization of linear interpolation
kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γ
nn(t), . . . , γ
n(k−1)n(t)
}, t ∈ [0, 1]
Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈
{nj, . . . ,n(j+ 1)
},
γmi (t) = (1− t) γm−1i
(t) + t γm−1i+1
(t), i ∈{nj, . . . ,n(j+ 1)−m
}
De Casteljau algorithm
Geometric interpretation
10
Bézier curves
Generalization of linear interpolation
kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γ
nn(t), . . . , γ
n(k−1)n(t)
}, t ∈ [0, 1]
Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈
{nj, . . . ,n(j+ 1)
},
γmi (t) = (1− t) γm−1i
(t) + t γm−1i+1
(t), i ∈{nj, . . . ,n(j+ 1)−m
}De Casteljau algorithm
Geometric interpretation
10
Bézier curves
Generalization of linear interpolation
kn+ 1 vertices {p0, . . . ,pkn} define k segments of degree n{γn0 (t), γ
nn(t), . . . , γ
n(k−1)n(t)
}, t ∈ [0, 1]
Defined recursively, for j ∈ {0, . . . , k− 1}γ0i (t) = pi, i ∈
{nj, . . . ,n(j+ 1)
},
γmi (t) = (1− t) γm−1i
(t) + t γm−1i+1
(t), i ∈{nj, . . . ,n(j+ 1)−m
}De Casteljau algorithm
Geometric interpretation
10
Bernstein polynomials
Algebraic interpretation
Expanding and collecting the pi terms,
γni (t) =n∑
j=0
(nj
)(1− t)n−j tj pi+j
Using Bernstein polynomials
γni (t) =n∑
j=0
bj,n(t)pi+j with bj,n(t) =(nj
)(1− t)n−j tj
Basis for the space of polynomials Pn with degree n or less (Why?)
11
Bernstein polynomials
Algebraic interpretation
Expanding and collecting the pi terms,
γni (t) =n∑
j=0
(nj
)(1− t)n−j tj pi+j
Using Bernstein polynomials
γni (t) =n∑
j=0
bj,n(t)pi+j with bj,n(t) =(nj
)(1− t)n−j tj
Basis for the space of polynomials Pn with degree n or less (Why?)
11
Bernstein polynomials
Algebraic interpretation
Expanding and collecting the pi terms,
γni (t) =n∑
j=0
(nj
)(1− t)n−j tj pi+j
Using Bernstein polynomials
γni (t) =n∑
j=0
bj,n(t)pi+j with bj,n(t) =(nj
)(1− t)n−j tj
Basis for the space of polynomials Pn with degree n or less (Why?)
11
Bernstein polynomials
Algebraic interpretation
Expanding and collecting the pi terms,
γni (t) =n∑
j=0
(nj
)(1− t)n−j tj pi+j
Using Bernstein polynomials
γni (t) =n∑
j=0
bj,n(t)pi+j with bj,n(t) =(nj
)(1− t)n−j tj
Basis for the space of polynomials Pn with degree n or less (Why?)
11
Control points and blending weights
In matrix form
γni (t) =[pni pni+1 · · · pni+n
]
︸ ︷︷ ︸Bézier control points CB
i
b0,n(t)
b1,n(t)...
bn,n(t)
blending weights Bn(t)
= CBi Bn(t)
Linear invariance is quite obvious in this form
12
Control points and blending weights
In matrix form
γni (t) =[pni pni+1 · · · pni+n
]︸ ︷︷ ︸
Bézier control points CBi
b0,n(t)
b1,n(t)...
bn,n(t)
blending weights Bn(t)
= CBi Bn(t)
Linear invariance is quite obvious in this form
12
Control points and blending weights
In matrix form
γni (t) =[pni pni+1 · · · pni+n
]︸ ︷︷ ︸
Bézier control points CBi
b0,n(t)
b1,n(t)...
bn,n(t)
blending weights Bn(t)
= CBi Bn(t)
Linear invariance is quite obvious in this form
12
Control points and blending weights
In matrix form
γni (t) =[pni pni+1 · · · pni+n
]︸ ︷︷ ︸
Bézier control points CBi
b0,n(t)
b1,n(t)...
bn,n(t)
blending weights Bn(t)
= CBi Bn(t)
Linear invariance is quite obvious in this form
12
Change of basis
Can be converted back and forth to power basis
Pn(t) =[1 t · · · tn
]TBn(t) = Bn Pn(t)
Examples
B1 =
[1 −1
0 1
]B2 =
1 −2 1
0 2 −2
0 0 1
B3 =
1 −3 3 −1
0 3 −6 3
0 0 3 −3
0 0 0 1
γni (t) = CBi Bn(t) = CBi
Change of basis︷︸︸︷Bn︸ ︷︷ ︸
Power basis control points CPi
Pn(t) = CPi Pn(t)
13
Change of basis
Can be converted back and forth to power basis
Pn(t) =[1 t · · · tn
]TBn(t) = Bn Pn(t)
Examples
B1 =
[1 −1
0 1
]B2 =
1 −2 1
0 2 −2
0 0 1
B3 =
1 −3 3 −1
0 3 −6 3
0 0 3 −3
0 0 0 1
γni (t) = CBi Bn(t) = CBi
Change of basis︷︸︸︷Bn︸ ︷︷ ︸
Power basis control points CPi
Pn(t) = CPi Pn(t)
13
Change of basis
Can be converted back and forth to power basis
Pn(t) =[1 t · · · tn
]TBn(t) = Bn Pn(t)
Examples
B1 =
[1 −1
0 1
]B2 =
1 −2 1
0 2 −2
0 0 1
B3 =
1 −3 3 −1
0 3 −6 3
0 0 3 −3
0 0 0 1
γni (t) = CBi Bn(t) = CBi
Change of basis︷︸︸︷Bn︸ ︷︷ ︸
Power basis control points CPi
Pn(t) = CPi Pn(t)
13
Bézier curves
Local control 3
Interpolates every nth point 3
Differentiable except (perhaps) at interpolation points 3
PostScript, SVG (Inkscape), RVG
Mathematica
14
Bézier curves
Local control 3
Interpolates every nth point 3
Differentiable except (perhaps) at interpolation points 3
PostScript, SVG (Inkscape), RVG
Mathematica
14
Bézier curves
Local control 3
Interpolates every nth point 3
Differentiable except (perhaps) at interpolation points 3
PostScript, SVG (Inkscape), RVG
Mathematica
14
Bézier curves
Local control 3
Interpolates every nth point 3
Differentiable except (perhaps) at interpolation points 3
PostScript, SVG (Inkscape), RVG
Mathematica
14
Bézier curves
Local control 3
Interpolates every nth point 3
Differentiable except (perhaps) at interpolation points 3
PostScript, SVG (Inkscape), RVG
Mathematica
14
Affine invariance of Bézier segments
Let T be an affine transformation and let
γn(t) =n∑
j=0
bj,n(t)pj
be a Bézier curve segment.
We want to show that
T( n∑
j=0
bj,n(t)pj
)=
n∑j=0
bj,n(t) T(pj).
This will be true if and only if all points in the Bézier curve are affine
combinations of the control points.
15
Affine invariance of Bézier segments
Let T be an affine transformation and let
γn(t) =n∑
j=0
bj,n(t)pj
be a Bézier curve segment.
We want to show that
T( n∑
j=0
bj,n(t)pj
)=
n∑j=0
bj,n(t) T(pj).
This will be true if and only if all points in the Bézier curve are affine
combinations of the control points.
15
Affine invariance of Bézier segments
Let T be an affine transformation and let
γn(t) =n∑
j=0
bj,n(t)pj
be a Bézier curve segment.
We want to show that
T( n∑
j=0
bj,n(t)pj
)=
n∑j=0
bj,n(t) T(pj).
This will be true if and only if all points in the Bézier curve are affine
combinations of the control points.
15
Affine invariance of Bézier segments
Indeed,n∑
j=0
bj,n(t) =n∑
j=0
(nj
)(1− t)n−j tj
=((1− t) + t
)n= 1n = 1.
The Bernstein polynomials therefore form a partition of unity
To apply an affine transformation to a Bézier curve, simply transform
the control points
16
Affine invariance of Bézier segments
Indeed,n∑
j=0
bj,n(t) =n∑
j=0
(nj
)(1− t)n−j tj =
((1− t) + t
)n
= 1n = 1.
The Bernstein polynomials therefore form a partition of unity
To apply an affine transformation to a Bézier curve, simply transform
the control points
16
Affine invariance of Bézier segments
Indeed,n∑
j=0
bj,n(t) =n∑
j=0
(nj
)(1− t)n−j tj =
((1− t) + t
)n= 1n = 1.
The Bernstein polynomials therefore form a partition of unity
To apply an affine transformation to a Bézier curve, simply transform
the control points
16
Affine invariance of Bézier segments
Indeed,n∑
j=0
bj,n(t) =n∑
j=0
(nj
)(1− t)n−j tj =
((1− t) + t
)n= 1n = 1.
The Bernstein polynomials therefore form a partition of unity
To apply an affine transformation to a Bézier curve, simply transform
the control points
16
Affine invariance of Bézier segments
Indeed,n∑
j=0
bj,n(t) =n∑
j=0
(nj
)(1− t)n−j tj =
((1− t) + t
)n= 1n = 1.
The Bernstein polynomials therefore form a partition of unity
To apply an affine transformation to a Bézier curve, simply transform
the control points
16
Convex hull property for Bézier segments
p =∑
i αipi is a convex combination of {pi} if∑
i αi = 1 and αi ≥ 0.
A set of points C is convex if every convex combination of points in C
also belongs to C
The convex hull of a set points S is the smallest convex set that
contains S
If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]
}is contained in the convex
hull of its control points
• From partition of unity and positivity in [0, 1]
• Useful for curve intersection, quick bounding box, etc
17
Convex hull property for Bézier segments
p =∑
i αipi is a convex combination of {pi} if∑
i αi = 1 and αi ≥ 0.
A set of points C is convex if every convex combination of points in C
also belongs to C
The convex hull of a set points S is the smallest convex set that
contains S
If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]
}is contained in the convex
hull of its control points
• From partition of unity and positivity in [0, 1]
• Useful for curve intersection, quick bounding box, etc
17
Convex hull property for Bézier segments
p =∑
i αipi is a convex combination of {pi} if∑
i αi = 1 and αi ≥ 0.
A set of points C is convex if every convex combination of points in C
also belongs to C
The convex hull of a set points S is the smallest convex set that
contains S
If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]
}is contained in the convex
hull of its control points
• From partition of unity and positivity in [0, 1]
• Useful for curve intersection, quick bounding box, etc
17
Convex hull property for Bézier segments
p =∑
i αipi is a convex combination of {pi} if∑
i αi = 1 and αi ≥ 0.
A set of points C is convex if every convex combination of points in C
also belongs to C
The convex hull of a set points S is the smallest convex set that
contains S
If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]
}is contained in the convex
hull of its control points
• From partition of unity and positivity in [0, 1]
• Useful for curve intersection, quick bounding box, etc
17
Convex hull property for Bézier segments
p =∑
i αipi is a convex combination of {pi} if∑
i αi = 1 and αi ≥ 0.
A set of points C is convex if every convex combination of points in C
also belongs to C
The convex hull of a set points S is the smallest convex set that
contains S
If γ is a Bézier curve, then{γ(t) | t ∈ [0, 1]
}is contained in the convex
hull of its control points
• From partition of unity and positivity in [0, 1]
• Useful for curve intersection, quick bounding box, etc
17
Derivative of Bézier segment
Since derivative operator is linear and γn(t) =∑n
j=0 bj,n(t)pj, all we
have to do is differentiate the Bernstein polynomials(bj,n(t)
)′=((
nj
)(1− t)n−j tj
)′
= j(nj
)(1− t)n−j tj−1 − (n− j)
(nj
)(1− t)n−1−j tj
= n(n−1j−1
)(1− t)(n−1)−(j−1) tj−1 − n
(n−1j
)(1− t)n−1−j tj
= n(bj−1,n−1(t)− bj,n−1(t)
)Therefore,
(γn)′(t) =n−1∑j=0
bj,n−1(t)qj with qj = n(pj+1 − pj).
18
Derivative of Bézier segment
Since derivative operator is linear and γn(t) =∑n
j=0 bj,n(t)pj, all we
have to do is differentiate the Bernstein polynomials(bj,n(t)
)′=((
nj
)(1− t)n−j tj
)′= j
(nj
)(1− t)n−j tj−1 − (n− j)
(nj
)(1− t)n−1−j tj
= n(n−1j−1
)(1− t)(n−1)−(j−1) tj−1 − n
(n−1j
)(1− t)n−1−j tj
= n(bj−1,n−1(t)− bj,n−1(t)
)Therefore,
(γn)′(t) =n−1∑j=0
bj,n−1(t)qj with qj = n(pj+1 − pj).
18
Derivative of Bézier segment
Since derivative operator is linear and γn(t) =∑n
j=0 bj,n(t)pj, all we
have to do is differentiate the Bernstein polynomials(bj,n(t)
)′=((
nj
)(1− t)n−j tj
)′= j
(nj
)(1− t)n−j tj−1 − (n− j)
(nj
)(1− t)n−1−j tj
= n(n−1j−1
)(1− t)(n−1)−(j−1) tj−1 − n
(n−1j
)(1− t)n−1−j tj
= n(bj−1,n−1(t)− bj,n−1(t)
)Therefore,
(γn)′(t) =n−1∑j=0
bj,n−1(t)qj with qj = n(pj+1 − pj).
18
Derivative of Bézier segment
Since derivative operator is linear and γn(t) =∑n
j=0 bj,n(t)pj, all we
have to do is differentiate the Bernstein polynomials(bj,n(t)
)′=((
nj
)(1− t)n−j tj
)′= j
(nj
)(1− t)n−j tj−1 − (n− j)
(nj
)(1− t)n−1−j tj
= n(n−1j−1
)(1− t)(n−1)−(j−1) tj−1 − n
(n−1j
)(1− t)n−1−j tj
= n(bj−1,n−1(t)− bj,n−1(t)
)
Therefore,
(γn)′(t) =n−1∑j=0
bj,n−1(t)qj with qj = n(pj+1 − pj).
18
Derivative of Bézier segment
Since derivative operator is linear and γn(t) =∑n
j=0 bj,n(t)pj, all we
have to do is differentiate the Bernstein polynomials(bj,n(t)
)′=((
nj
)(1− t)n−j tj
)′= j
(nj
)(1− t)n−j tj−1 − (n− j)
(nj
)(1− t)n−1−j tj
= n(n−1j−1
)(1− t)(n−1)−(j−1) tj−1 − n
(n−1j
)(1− t)n−1−j tj
= n(bj−1,n−1(t)− bj,n−1(t)
)Therefore,
(γn)′(t) =n−1∑j=0
bj,n−1(t)qj with qj = n(pj+1 − pj).
18
Derivative of Bézier segment
What is the derivative at the endpoints?
How do we connect segments so that they are C1 continuous?
What about G1 continuity?
Show in Inkscape
19
Derivative of Bézier segment
What is the derivative at the endpoints?
How do we connect segments so that they are C1 continuous?
What about G1 continuity?
Show in Inkscape
19
Derivative of Bézier segment
What is the derivative at the endpoints?
How do we connect segments so that they are C1 continuous?
What about G1 continuity?
Show in Inkscape
19
Derivative of Bézier segment
What is the derivative at the endpoints?
How do we connect segments so that they are C1 continuous?
What about G1 continuity?
Show in Inkscape
19
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t)
=(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj
= n+1n+1−j
(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj
= n+1j+1
t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi
20
Degree elevation of Bézier segment
Express a segment γn(t) as γn+1(t)? (write bj,n(t) in terms of bk,n+1(t)?)
Easy to express both bj,n+1(t) and bj+1,n+1(t) in terms of bj,n(t)
bj,n+1(t) =(n+1j
)(1− t)n+1−j tj = n+1
n+1−j(1− t)bj,n(t)
bj+1,n+1(t) =(n+1j+1
)(1− t)n−j tj = n+1
j+1t bj,n(t)
From these,
bj,n(t) =n+1−jn+1 bj,n+1(t) +
j+1n+1 bj+1,n+1(t)
Expanding and collecting terms,
γn(t) =n∑
i=0
bi,n(t)pi =n+1∑j=0
bj,n+1(t)qj = γn+1(t)
with q0 = p0, qn+1 = pn, and
qi =j
n+1 pi−1 + (1− jn+1)pi 20
Degree elevation of Bézier segment
Examples [p0 p1
]⇔
[p0
12(p0 + p1) p1
][p0 p1 p2
]⇔
[p0
13(p0 + 2p1)
13(2p1 + p2) p2
]
21
References
G. Farin. Curves and Surfaces for CAGD, A Practical Guide, 5th edition.
Morgan Kaufmann, 2002.
22
